(x^2+13x)=(6x+18)

Solution for (x^2+13x)=(6x+18) equation:

(x^2+13x)=(6x+18)

We move all terms to the left:

(x^2+13x)-((6x+18))=0

We get rid of parentheses

x^2+13x-((6x+18))=0


We calculate terms in parentheses: -((6x+18)), so:

(6x+18)

We get rid of parentheses

6x+18

Back to the equation:

-(6x+18)


We get rid of parentheses

x^2+13x-6x-18=0

We add all the numbers together, and all the variables

x^2+7x-18=0

a = 1; b = 7; c = -18;
Δ = b2-4ac
Δ = 72-4·1·(-18)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-11}{2*1}=\frac{-18}{2}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+11}{2*1}=\frac{4}{2}
=2
$